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Control Systems Engineering : Masons Gain

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Here is the system I am looking at

    image.jpg

    2. Relevant equations

    So I worked out a formula for the steady state error and put it in a form similar to masons gain formula

    image.jpg

    3. The attempt at a solution

    But when I try and draw a block diagram and work out e I get confused by forward path and loop gain. It's not a question I have been asked, I am just playing, perhaps I cannot apply masons gain to my drawing? Or my drawing is wrong?

    image.jpg
     
  2. jcsd
  3. Jan 28, 2013 #2

    rude man

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    Your equations are not correct. For one thing, they are not dimensionally consistent: dh/dt = qin - qout has different dimensions on each side. q has dimension L^3 whereas dh/dt has dimension LT^(-1). L = length, T = time.

    Your block diagram is good up to the point qin but then falls apart. Correct your equation for dh/dt: h cannot = qin - qout, you've already written dh/dt = qin - qout (which was incorrect also).

    What is the correct relationship between dh/dt and qin - qout? (Hint: area of the bottom of the tank is a factor). And, once you get dh/dt right, how does one go from dh/dt to h?
     
  4. Jan 28, 2013 #3
    Thanks for pointing that out, so I came to the idea that

    a*dh/dt = dv/dt = q_in(t) - q_out(t) = 0 for steady state

    which would make the rest of the derivation ok I hope?

    Which then gave me a different drawing for the block diagram.

    image.jpg

    Which gives me a loop gain of (c_1 - c_2)/as

    Instead of the c_1 / c_2 i got when I used the assumption that dv/dt = 0 for steady state. Am I still wrong with my equations or drawing?

    I thought I had it for a minute there! Thanks for your help. I don't need to do it, it's just bugging me as I cannot relate it to the rest of my notes.
     
  5. Jan 28, 2013 #4

    rude man

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    Your block diagram is now correct.

    Loop gain = H(s)/H0(s).
     
  6. Jan 29, 2013 #5
    Thanks again mate, I played with it some more, got the answers I wanted and have a much better understanding of signal flow graphs and masons formula.
     
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